Memorandum from R. L. Murray to Dr. Clifford K. Beck
Typescript
2 pp.
July 5, 1951
MurNBeffect070551


[ OFFICIAL USE ONLY]

Murray
July 5, 1951
NCSC-23

TO: Clifford K. Beck
FROM: R. L. Murray
COPIES TO: Reactor Committee
SUBJECT: Effect of Pb and Bismuth in Thermal Column.

Information was received from Los Alamos that there was a factor of 10 dif-
ference
in neutron flux at the end of the thermal column between the cases
of lead and bismuth gamma shielding. The shield thickness used there is
8"; our calculations had indicated that 4" of Pb was adequate, for which it
had been thought that the neutron attenuation was not serious.

As a check on the Los Alamos quotations however, the flux values are computed,
for an 8" spherical shell, as sketched below.

Analysis: the flux equations for the carbon and shield are

where K takes on the value Kc in carbon, Ks in the shield

In the three regions, solutions are assumed



Where the thermal column is assumed to be infinite in extent for calculation
purposes.

Boundary conditions are taken as follows:
[phi]I = 1 at the surface of the core. Normalization to
any known value can be made later.


[page 2]


A comparison between the coefficients E for lead and bismuth serves to de-
termine
the relative attenuation, since the shape of the flux curve will be
the same in the two cases.

By the application of standard, methods of solution (Soodak and Campbell)
including calculation of the K's (reciprocals of the thermal neutron diffusion
lengths) and the [lambda]'s (transport mean free paths), the coefficient E is
computed.

The general solution is:
where =
, and

The constants used were:
Kc = 0.020 cm-1 ; [lambda]c = 2.72 cm r1 = 66 cm
Kpb = 0.085 cm-1 ; [lambda]Pb = 2.33 cm r2 = 86.3 cm
KBi = 0.0187 cm-1; [lambda]Bi = 3.85 cm
For the case of 8" of lead, E is 0.185; for 8" of bismuth E is approximately
1, (which is expected because the diffusion lengths for C and Bi are about
the same). Thus the flux with a lead shield is about 1/5 that with a bismuth
shield. No account is taken of impurities in either case. A separate
calculation with a 4" lead shield leads to a factor of 1/3, which would not
be too serious a drop. The problem thus rests on the necessary shield
thickness for protection against gammas.